Question:

A 2 kg object is initially at rest on a frictionless surface. It is then subjected to two forces: a horizontal force of 12 N applied to the right and a vertical force of 15 N applied upwards. The object moves a distance of 3 meters horizontally. Calculate the work done by each force and the net work done on the object.

Solution:

Given: Mass of the object, m = 2 kg Horizontal force, F₁ = 12 N Vertical force, F₂ = 15 N Distance moved, d = 3 m

To calculate the work done by each force, we'll use the formula:

Work done, W = Force × Displacement × cos(θ)

where θ is the angle between the direction of the force and the direction of displacement.

1. Work done by the horizontal force:

The horizontal force is applied in the same direction as the displacement, so θ = 0°.

W₁ = F₁ × d × cos(0°)

Since cos(0°) = 1, we have:

W₁ = F₁ × d

Substituting the values:

W₁ = 12 N × 3 m = 36 J

So, the work done by the horizontal force is 36 J.

2. Work done by the vertical force:

The vertical force is applied perpendicular to the displacement, so θ = 90°.

W₂ = F₂ × d × cos(90°)

Since cos(90°) = 0, the work done by the vertical force is zero.

3. Net work done on the object:

The net work done on the object is the sum of the work done by each force.

Net work = W₁ + W₂ = 36 J + 0 J = 36 J

Therefore, the net work done on the object is 36 J.